Commit 5cf84547 authored by Olivier's avatar Olivier

Exercises 3: proof-read

* refreshed formula sheet
* labeled non-examinable exercises
* fixed page layout somewhat
parent 5e43b11d
\renewcommand{\lastedityear}{2019}
\renewcommand{\lasteditmonth}{03}
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\atstartofexercises
\fluidmechexercisestitle
......@@ -10,35 +10,30 @@
\mecafluexboxen
\begin{boiboiboite}
Reynolds Transport Theorem:
\begin{IEEEeqnarray}{rCl}
\timederivative{B_\sys} & = & \timederivative{} \iiint_\cv \rho b \diff \vol + \iint_\cs \rho b \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt}}
\end{IEEEeqnarray}
Mass conservation:
\begin{IEEEeqnarray}{rCCCCCl}
\timederivative{m_\sys} & = & 0 & = & \timederivative{} \iiint_\cv \rho \diff \vol & + & \iint_\cs \rho \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_mass}}
\end{IEEEeqnarray}
Change in linear momentum:
\begin{IEEEeqnarray}{rCCCCCl}
\timederivative{(m \vec V_{sys})} & = & \vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol & + & \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_linearmom}}
\end{IEEEeqnarray}
Change in angular momentum:
\begin{equation}
\timederivative{(\vec r_{\X m} \wedge m \vec V)_\sys} = \vec M_{\net, \X} = \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol + \iint_\cs \vec r_{\X m} \wedge \rho \ (\vec V_\rel \cdot \vec n) \vec V \diff A \tag{\ref{eq_rtt_angularmom}}
\end{equation}
Mass balance through an arbitrary volume:
\begin{IEEEeqnarray}{rCCCl}
0 & = & \timederivative{} \iiint_\cv \rho \diff \vol & + & \iint_\cs \rho \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_mass}}
\end{IEEEeqnarray}
Momentum balance through an arbitrary volume:
\begin{IEEEeqnarray}{rCCCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol & + & \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_linearmom}}
\end{IEEEeqnarray}
Angular momentum balance through an arbitrary volume:
\begin{IEEEeqnarray}{rCCCl}
\vec M_{\net, \X} &=& \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol &+& \iint_\cs \vec r_{\X m} \wedge \rho \ (\vec V_\rel \cdot \vec n) \vec V \diff A \ztag{\ref{eq_rtt_angularmom}}
\end{IEEEeqnarray}
\end{boiboiboite}
\clearpage%handmade
%%%%
\subsubsection{Pipe bend}
\wherefrom{\cczero \oc}
\label{exo_pipe_bend}
A pipe with diameter~\SI{30}{\milli\meter} has a bend with angle $\theta = \SI{130}{\degree}$, as shown in \cref{fig_pipe_bend}. Water enters and leaves the pipe with the same speed $V_1 = V_2 = \SI{1,5}{\metre\per\second}$. The velocity distribution at both inlet and outlet is uniform.
\begin{figure}
\begin{figure}[ht!]
\begin{center}
\includegraphics[width=7.5cm]{pipe_bend}
\vspace{-0.5cm}
......@@ -53,9 +48,6 @@ Change in angular momentum:
\item What would be the new force if all of the speeds were doubled?
\end{enumerate}
%%%%
\subsubsection{Exhaust gas deflector}
\label{exo_exhaust_gas_deflector}
......@@ -86,8 +78,7 @@ Change in angular momentum:
\label{exo_pelton_turbine}
A water turbine is modeled as the following system: a water jet exiting a stationary nozzle hits a blade which is mounted on a rotor (\cref{fig_water_turbine}). In the ideal case, viscous effects can be neglected, and the water jet is deflected entirely with a~\SI{180}{\degree} angle.
\begin{figure}
\begin{figure}[ht!]
\begin{center}
\includegraphics[width=12cm]{water_turbine_blade}
\end{center}
......@@ -258,6 +249,7 @@ Change in angular momentum:
%%%%
\subsubsection{Moment on gas deflector}
\label{exo_moment_gas_deflector}
\wherefrom{non-examniable}
We revisit the exhaust gas deflector of exercise \ref{exo_exhaust_gas_deflector} p.\pageref{exo_exhaust_gas_deflector}. \Cref{fig_deflector_sideview} below shows the deflector viewed from the side. The midpoint of the inlet is \SI{2}{\metre} above and \SI{5}{\metre} behind the wheel labeled~“\textbf{A}”, while the midpoint of the outlet is \SI{3,5}{\metre} above and \SI{1,5}{\metre} behind it.
\begin{figure}[ht!]
......@@ -270,10 +262,11 @@ Change in angular momentum:
What is the moment generated by the gas flow about the axis of the wheel labeled “\textbf{A}”?
\clearpage%handmade
%%%%
\subsubsection{Helicopter tail moment}
\label{exo_helicopter_tail_moment}
\wherefrom{non-examinable}
In a helicopter, the role of the tail is to counter exactly the moment exerted by the main rotor about the main rotor axis. This is usually done using a tail rotor which is rotating around a horizontal axis.
......@@ -297,7 +290,7 @@ Change in angular momentum:
\item Propose and quantify a modification to the tail geometry or operating conditions that would allow the tail to produce no thrust (that is to say, zero force in the $x$-axis), while still generating the same moment.
\end{enumerate}
\textit{Remark: this system is commercialized by MD Helicopters as the \wed{NOTAR}{\textsc{notar}}. The use of exhaust gases was abandoned, however, a clever use of air circulation around the tail pipe axis contributes to the generated moment; this effect is explored in chapter~8 (\S\ref{ch_circulating_cylinder} p.\pageref{ch_circulating_cylinder}).}
\textit{Remark: this system is commercialized by MD Helicopters as the \wed{NOTAR}{\textsc{notar}}. The use of exhaust gases was abandoned, however, a clever use of air circulation around the tail pipe axis contributes to the generated moment; this effect is explored in \chaptereleven (\S\ref{ch_circulating_cylinder} p.\pageref{ch_circulating_cylinder}).}
%%%%
......@@ -407,7 +400,7 @@ Change in angular momentum:
\tab $V_\text{center} = \num{1,2245} U$; $F_\net = \SI{+393}{\newton}$ (positive!)
\item [\ref{exo_thrust_reverser}]%
\tab $\dot m_\text{cold} = \SI{297}{\kilogram\per\second}$, $\dot m_\text{hot} = \SI{59,4}{\kilogram\per\second}$;
\tab $F_\text{cold flow, normal, bench \& runway} = \SI{+74,25}{\kilo\newton}$, $F_\text{hot flow, normal \& reverse, bench \& runway} = \SI{+8,316}{\kilo\newton}$;
\tab $F_\text{cold flow, normal, bench \& runway} = \SI{+74,25}{\kilo\newton}$,\\ $F_\text{hot flow, normal \& reverse, bench \& runway} = \SI{+8,316}{\kilo\newton}$;
\tab $F_\text{cold flow, reverse, bench \& runway} = \SI{-24,35}{\kilo\newton}$ and $F_\text{hot flow, normal, bench \& runway} = \SI{+8,316}{\kilo\newton}$.
\tab Adding the net pressure force due to the (lossless) flow acceleration upstream of the inlet, we obtain, on the bench: $F_\text{engine bench, normal} = \SI{-93,26}{\kilo\newton}$, $F_\text{engine bench, reverse} = \SI{+5,344}{\kilo\newton}$;
and on the runway: $F_\text{engine runway, normal} = \SI{-92,5}{\kilo\newton}$ and $F_\text{engine bench, reverse} = \SI{+6,094}{\kilo\newton}$.
......
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